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Proof step 5 from proof equal-minus-one (previous step)

i = ~1, i ^ j = ~1, i ≠ 0 ∨ 0 < j ⊦ j = 0, abs(j) % 2 = 1 ∧ i = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
~1 ^ j = ~1 ⊦ j = 0, abs(j) % 2 = 1

Rule argument

Additional information